使用PHP编辑SQL数据库
[英]Editing SQL Database using PHP
问题描述
在使用下面给出的PHP代码编辑特定记录时，同时将数据库中的所有记录编辑为某些垃圾值。 这里的“ db”是数据库。 我是PHP和SQL的新手。 请帮忙
<?php
/*
EDIT.PHP
Allows user to edit specific entry in database
*/
// creates the edit record form
// since this form is used multiple times in this file, I have made it a function that is easily reusable
function renderForm($reportno, $dateofreceipt, $title, $type, $issuingagency, $markedto, $date, $remarks, $isdate, $issuedto, $returndate)
{
?>
<!DOCTYPE HTML PUBLIC >
<html>
<head>
<title>Edit Record</title>
</head>
<body>
<form action="edit.php" method="post">
<div>
<p><strong>Report No.:</strong> <?php echo $reportno; ?></p>
<strong>Date of receipt: *</strong> <input type="date" name="dateofreceipt" value="<?php echo $dateofreceipt; ?>"/><br/>
<strong>Report Title: *</strong> <input type="text" name="title" value="<?php echo $title; ?>"/><br/>
<strong>Report Type: *</strong> <input type="text" name="type" value="<?php echo $type; ?>"/><br/>
<strong>Issuing agency: *</strong> <input type="text" name="issuingagency" value="<?php echo $issuingagency; ?>"/><br/>
<strong>Marked to: *</strong> <input type="text" name="markedto" value="<?php echo $markedto; ?>"/><br/>
<strong>Date: *</strong> <input type="date" name="date" value="<?php echo $date; ?>"/><br/>
<strong>Remarks: *</strong> <input type="text" name="remarks" value="<?php echo $remarks; ?>"/><br/>
<strong>Issuing Date: *</strong> <input type="date" name="isdate" value="<?php echo $isdate; ?>"/><br/>
<strong>Issued To: *</strong> <input type="text" name="issuedto" value="<?php echo $issuedto; ?>"/><br/>
<strong>Return Date: *</strong> <input type="date" name="returndate" value="<?php echo $returndate; ?>"/><br/>
<p>* Required</p>
<input type="submit" name="submit" value="Submit">
</div>
</form>
</body>
</html>
<?php
}
// connect to the database
include('connect-db.php');
// check if the form has been submitted. If it has, process the form and save it to the database
if (isset($_POST['submit']))
{
// get form data, making sure it is valid
$reportno = $_POST['reportno'];
$dateofreceipt = mysql_real_escape_string(htmlspecialchars($_POST['dateofreceipt']));
$title = mysql_real_escape_string(htmlspecialchars($_POST['title']));
$type = mysql_real_escape_string(htmlspecialchars($_POST['type']));
$issuingagency = mysql_real_escape_string(htmlspecialchars($_POST['issuingagency']));
$markedto = mysql_real_escape_string(htmlspecialchars($_POST['markedto']));
$date = mysql_real_escape_string(htmlspecialchars($_POST['date']));
$remarks = mysql_real_escape_string(htmlspecialchars($_POST['remarks']));
$isdate = mysql_real_escape_string(htmlspecialchars($_POST['isdate']));
$issuedto = mysql_real_escape_string(htmlspecialchars($_POST['issuedto']));
$returndate = mysql_real_escape_string(htmlspecialchars($_POST['returndate']));
//renderForm($reportno, $dateofreceipt, $title, $type, $issuingagency, $markedto, $date,$remarks, $isdate, $issuedto, $returndate, $error);
// save the data to the database
mysql_query("UPDATE `db` SET `Report No.`='[$reportno]',`Date of receipt`='[$dateofreceipt]',`Report Title`='[$title]',`Report Type`='[$type]',`Issuing agency`='[$issuingagency]',`Marked to`='[$markedto]',`Date`='[$date]',`Remarks`='[$remarks]',`Issuing date`='[$isdate]',`Issued to`='[$issuedto]',`Return Date`='[$returndate]' WHERE `Report No.`= '$id'")
// once saved, redirect back to the view page
header("Location: view.php");
}
// query db
$id = $_GET['id'];
$result = mysql_query("SELECT * FROM db WHERE `Report No.`= '$id'")
or die(mysql_error());
$row = mysql_fetch_array($result);
// check that the 'id' matches up with a row in the databse
if($row)
{
// get data from db
$reportno = $row['Report No.'];
$dateofreceipt = $row['Date of receipt'];
$title= $row['Report Title'];
$type= $row['Report Type'];
$issuingagency= $row['Issuing agency'];
$markedto= $row['Marked to'];
$date= $row['Date'];
$remarks=$row['Remarks'];
$isdate= $row['Issuing date'];
$issuedto= $row['Issued to'];
$returndate= $row['Return Date'];
// show form
renderForm($reportno, $dateofreceipt, $title, $type, $issuingagency, $markedto, $date, $remarks ,$isdate, $issuedto, $returndate, '');
}
?>
3 个解决方案
解决方案1
0 2016-06-20 07:20:00
这里有几个问题：
已弃用PhP中的mysql api。 不要打赌它会工作更长的时间。 请改用mysqli api。
在查询中，“
where 1部分完全多余。1表示true，where 1表示所有记录，这时您可以完全删除WHERE。您可能想使用WHERE somekey = 1，这是不同的。
解决方案2
0 已采纳 2016-06-20 07:24:59
尝试使用此查询进行更新。 也不要忘记在语句后添加分号。 使用mysqli_ *函数代替mysql_ *
mysqli_query("UPDATE `db` SET `Date of receipt`='$dateofreceipt',`Report Title`='$title',`Report Type`='$type',`Issuing agency`='$issuingagency',`Marked to`='$markedto',`Date`='$date',`Remarks`='$remarks',`Issuing date`='$isdate',`Issued to`='$issuedto',`Return Date`='$returndate' WHERE Report No = $reportno");
解决方案3
0 2016-06-20 07:54:15
尝试这个
mysql_query（“ UPDATE db SET Report No. =”。'$ reportno'。“， Date of receipt =。”'$ dateofreceipt'。“， Report Title =。”'$ title'。“， Report Type =。”' $ type'。“， Issuing agency =。”'$ issuingagency'。“， Marked to =。”'$ markedto'。“， Date =。”'$ date'。“， Remarks =。”'$ remarks'。 “， Issuing date =。”'$ isdate'。“， Issued to =。”'$ issuedto'。“， Return Date =。”'$ returndate'。“ Report No. =。”'$ id'。“ ”）
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